How do you simplify #(c^3d^4)/(cd^7)#?

2 Answers
Aug 31, 2017

Answer:

See a solution process below:

Explanation:

First, rewrite the expression as:

#(c^3/c)(d^4/d^7)#

Next, use these rules of exponents to simplify the #c# terms:

#a = a^color(blue)(1)# and #x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))#

#(c^3/c^color(blue)(1))(d^4/d^7) => (c^color(red)(3)/c^color(blue)(1))(d^4/d^7) => c^(color(red)(3)-color(blue)(1))(d^4/d^7) => c^2(d^4/d^7)#

Now, use this rule of exponents to simplify the #d# terms:

#x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))#

#c^2(d^color(red)(4)/d^color(blue)(7)) => c^2(1/d^(color(blue)(7)-color(red)(4))) => c^2 * 1/d^3 = c^2/d^3#

Aug 31, 2017

Answer:

#(c^2)/(d^3)#

Explanation:

#"using the "color(blue)"law of exponents"#

#•color(white)(x)a^m/a^n=a^(m-n)to"for "m>n#

#•color(white)(x)a^m/a^n=1/a^(n-m)to" for "n>m#

#(c^3d^4)/(cd^7)#

#=c^3/c^1xxd^4/d^7#

#=c^((3-1))xx1/d^((7-4))#

#=c^2/1xx1/d^3=c^2/d^3#